नयाँ पाठ्यक्रम · New Curriculum 2083 · Model Question
ऐच्छिक गणित — कक्षा ९ र १०
∫ Optional Mathematics
⏱ ३ घण्टा / 3 Hours📝 ७५ पूर्णाङ्क / 75 Marks📋 17 Questions (3 Groups)✅ Full Answers + Marking Scheme
🆕 New Curriculum 2082Algebra · Trigonometry · GeometryVector · Statistics · LimitWCA + CCA QuestionsBilingual NP/EN
📌 नयाँ पाठ्यक्रम सूचना / New Curriculum Note: यो नमुना प्रश्नपत्र CDC ले प्रकाशित गरेको २०८२ को विशिष्टीकरण तालिका (Specification Grid) अनुसार तयार गरिएको हो। प्रश्नपत्रमा तीन समूह छन्: Group A (11 Objective), Group B (12 WCA Sub-questions = 40 marks), र Group C (4 CCA Questions = 24 marks)। New format: Objective (11×1=11) + Within Content Area WCA (40 marks) + Cross Content Area CCA (24 marks) = 75 marks total. You can download the original Specification Grid, Model Questions and Marking Scheme here.
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समूह 'क' — वस्तुगत प्रश्न (Group A — Objective Questions)
MCQ — Click to check your answer
[11×1 = 11]
A
लेखाचित्रको प्रयोगबाट ax² + bx + c = 0 लाई समाधान गर्दा वास्तविक मूलहरू …… सँग सम्बन्धित हुन्छन् ।
When solving ax² + bx + c = 0 graphically, the real roots correspond to ……
1 mark
a The vertex of the parabola
b The y-intercept where the parabola crosses the Y-axis
c The x-intercepts where the parabola crosses the X-axis
d The axis of symmetry
✓ Answer
c. The x-intercepts — where the parabola crosses the X-axis, y=0, so ax²+bx+c=0 at those points — those x values are the real roots.
B
यदि फलन f(x) = 3x – 2 छ भने तलका मध्ये कुन सत्य हुन्छ ?
If function f(x) = 3x – 2, which one of the following is true?
1 mark
a f⁻¹(3) = 7
b f⁻¹(7) = 3
c f(7) = 3
d f⁻¹(f⁻¹(3)) = 7
✓ Answer
b. f⁻¹(7) = 3 — since f(x) = 3x−2, set y=3x−2, so x=(y+2)/3. Thus f⁻¹(7)=(7+2)/3=9/3=3. ✓
C
तलका मध्ये cos 20° + cos 40° को मान कुन हो ?
Which of the following is the value of cos 20° + cos 40°?
1 mark
a 2 cos 30° cos 10°
b 2 sin 30° cos 30°
c 2 sin 30°
d 1
✓ Answer
a. 2 cos 30° cos 10° — using sum-to-product: cos C + cos D = 2 cos((C+D)/2) cos((C−D)/2). So cos 20° + cos 40° = 2 cos 30° cos 10°.
D
तलका मध्ये कुन अभिव्यञ्जकले sin(A + B) लाई सही रूपमा जनाउँछ ?
Which of the following expression correctly expresses sin(A + B)?
1 mark
a sin A cos B − cos A sin B
b sin A cos B + cos A sin B
c cos A cos B − sin A sin B
d sin A + sin B
✓ Answer
b. sin(A+B) = sin A cos B + cos A sin B — the standard compound angle formula for sine.
E
एकजना अवलोकनकर्ताले एउटै समतल सतहमा रहेको एउटा अग्लो रूखको टुप्पोलाई हेर्दा उन्नतांश कोणमा कस्तो परिवर्तन आउँछ जब उनी रूखको नजिक जान्छन् ?
What kind of change occurs in the angle of elevation when an observer moves closer to a tall tree on a plane surface?
1 mark
aदूरी घट्दै जाँदा उन्नतांश कोण घट्छ । The angle of elevation decreases.
bउन्नतांश कोण समान रहन्छ । The angle of elevation stays the same.
cरूखको उचाइ यथावत रही दूरी घट्दा उन्नतांश बढ्छ । As height stays constant and distance decreases, the angle of elevation increases.
dउचाइ यथावत हुँदा उन्नतांश कोण घट्छ । When height remains unchanged, the angle decreases.
✓ Answer
c. As the observer moves closer (distance decreases) while the tree height remains constant, tan θ = h/d increases, so the angle of elevation increases.
F
समीकरण x² + y² = 49 ले एउटा वृत्तको प्रतिनिधित्व गर्छ । यस वृत्तको व्यास कति हुन्छ ?
The equation x² + y² = 49 represents a circle. What is the diameter of the circle?
1 mark
a 7
b 14
c 49
d 98
✓ Answer
b. 14 — x²+y²=r² so r²=49, r=7. Diameter = 2r = 2×7 = 14.
G
दुई रेखाहरू एकआपसमा लम्ब छन् । यदि एउटा रेखाको झुकाव m भए अर्को रेखाको झुकाव ……. हुन नै पर्छ ।
Two lines are perpendicular. If one line has slope m, the slope of the other must be ……
1 mark
a m
b −m
c 1/m
d −1/m
✓ Answer
d. −1/m — for two perpendicular lines, the product of their slopes equals −1: m₁ × m₂ = −1, so m₂ = −1/m.
H
रेखाहरू y = x र x = 0 मा क्रमिक रूपमा गरिएको दुईओटा परावर्तन गरिएको छ । उक्त संयुक्त स्थानान्तरणमा कुन बिन्दुको प्रतिबिम्ब उही रहन्छ ?
Two reflections are performed successively in lines y = x and x = 0. Which point is invariant under the combined transformation?
1 mark
a Every point on y = x
b Every point on x = 0
c Only the origin (0, 0)
d Such a point does not exist.
✓ Answer
c. Only the origin — the origin (0,0) is the only point that lies on both lines y=x and x=0 simultaneously, so both reflections leave it unchanged.
I
यदि दुई non-zero vectors को स्केलर गुणनफल शून्य छ भने तिनीहरूबिचको कोणबारे के निष्कर्ष निकाल्न सकिन्छ ?
If the scalar product of two non-zero vectors is zero, what can be concluded about the angle between them?
1 mark
a −1
b 1
c 0 (vectors are perpendicular)
d k (k ≠ 0)
✓ Answer
c. 0 — a·b = |a||b|cosθ = 0 means cosθ = 0, so θ = 90°. The vectors are perpendicular.
J
एउटा फलन f(x) को x = a मा बायाँ पक्षको र दायाँ पक्षको सीमान्तमान बराबर छ तर f(a) परिभाषित छैन । यो फलनलाई x = a मा के भन्न सकिन्छ ?
A function f(x) has equal left-hand and right-hand limits at x = a, but f(a) is not defined. How should this function be described at x = a?
1 mark
a Continuous, because the left and right limits are equal.
b Continuous, because limits are more important than function values.
c Discontinuous, because the function has no limit at x = a.
d Discontinuous, because continuity requires the functional value, LHL and RHL should all be equal.
✓ Answer
d. Discontinuous — for continuity at x=a: (1) f(a) must be defined, (2) limit must exist, (3) limit must equal f(a). Since f(a) is not defined, condition 1 fails.
K
दुई थरी तथ्याङ्कको तुलना गर्न विचरणशीलताको गुणाङ्क प्रयोग गरिन्छ । उच्च विचरणशीलताले के जनाउँछ ?
The Coefficient of Variation (CV) is used to compare two datasets. What does a higher CV indicate?
1 mark
a More consistent data
b Greater variability / less consistency
c Lower mean
d Higher median
✓ Answer
b. Greater variability — a higher CV (= σ/x̄ × 100%) means data is more spread out and less consistent. Lower CV = more uniform data.
समूह 'ख' — विषयक्षेत्रगत प्रश्न (Group B — Within Content Area)
Questions 2–13 covering all 6 content areas
[40 marks]
2
बीजगिणत / Algebra
एउटा फलन f(x) = x – 1 र एउटा बहुपदीय अभिव्यञ्जक p(x) = x³ – 6x² + 11x – 6 दिइएका छन् ।
The function f(x) = x – 1 and polynomial p(x) = x³ – 6x² + 11x – 6 are given.
a. p(x) को एउटा गुणनखण्ड f(x) हुने अवस्था लेख्नुहोस् । b. बहुपदीय अभिव्यञ्जक p(x) का सम्भावित मूलहरू निर्धारण गर्न आनुपातिक मूल साध्य प्रयोग गर्नुहोस् ।
1+1
✓ Model Answer
a. Condition (1 mark): For (x–1) to be a factor of p(x), the remainder when p(x) is divided by (x–1) must be zero, i.e. p(1) = 0.
Verify: p(1) = 1 – 6 + 11 – 6 = 0 ✓ So (x–1) IS a factor.
b. Rational Root Theorem (1 mark):
Possible rational roots = ± (factors of constant term) / (factors of leading coefficient)
= ± (factors of 6) / (factors of 1) = ±1, ±2, ±3, ±6
Testing: p(1)=0 ✓, p(2)=8–24+22–6=0 ✓, p(3)=27–54+33–6=0 ✓
So the roots are x = 1, 2, 3
3
बीजगिणत / Algebra
फलन f(x) = x + 1 र g(x) = x² – 5x + 6 दिइएका छन् ।
Given f(x) = x + 1 and g(x) = x² – 5x + 6.
a. संयुक्त फलन f(g(x)) पत्ता लगाउनुहोस् । b. g(x) लाई y = (x – h)² + k को रूपमा लेख्नुहोस् । c. y = x² लाई g(x) मा स्थानान्तरण गर्न गरिएको संयुक्त स्थानान्तरण वर्णन गर्नुहोस् ।
b. Vertex form (1 mark):
g(x) = x² – 5x + 6 = (x – 5/2)² – 25/4 + 6 = (x – 5/2)² – 1/4
So h = 5/2, k = –1/4 g(x) = (x – 5/2)² – 1/4
c. Combined Transformation (2 marks):
y = x² → g(x) = (x – 5/2)² – 1/4 involves: Step 1: Translation of 5/2 units in the positive x-direction (right shift) Step 2: Translation of 1/4 units in the negative y-direction (downward shift)
4
बीजगिणत / Algebra — Matrix & Linear Programming
मेट्रिक्सहरू A = [[2,1],[1,3]] र B = [[x],[y]] दिइएका छन् ।
Matrices A = [[2,1],[1,3]] and B = [[x],[y]] are given.
a. Aᵀ पत्ता लगाउनुहोस् । b. A⁻¹ पत्ता लगाउनुहोस् । c. असमानताहरू 2x + y ≤ 8, x + 3y ≤ 9, x ≥ 0 र y ≥ 0 को साझा हल क्षेत्र लेखाचित्रमा देखाउनुहोस् ।
1+1+2
✓ Model Answer
a. Aᵀ (1 mark):
Aᵀ = [[2, 1], [1, 3]] — rows and columns interchanged: Aᵀ = [[2, 1], [1, 3]] (symmetric matrix — same as A)
c. Feasible Region (2 marks):
Boundary lines: 2x+y=8 (passes through (4,0) and (0,8)) and x+3y=9 (passes through (9,0) and (0,3)).
Intersection of the two lines: 2x+y=8 and x+3y=9 → solving: x=3, y=2 → Corner point (3,2).
Corner points of feasible region: (0,0), (4,0), (3,2), (0,3)
Shade the region satisfying all inequalities (below both lines, in first quadrant).
5
त्रिकोणमिति / Trigonometry
सर्वसमिका sin(A+B) = sinA cosB + cosA sinB दिइएको छ ।
Given the identity sin(A+B) = sin A cos B + cos A sin B.
a. दिइएको सर्वसमिकालाई sin 2A = 2 sin A cos A को स्वरूपमा कसरी परिवर्तन गर्न सकिन्छ ? b. यदि sin 2A = 2 sin A cos A भए sin 2A = 2tanA/(1+tan²A) प्रमाणित गर्नुहोस् ।
1+1
✓ Model Answer
a. Converting to sin 2A (1 mark):
Set B = A in the given identity:
sin(A+A) = sin A cos A + cos A sin A sin 2A = 2 sin A cos A ✓
b. Proving sin 2A = 2tanA/(1+tan²A) (1 mark):
L.H.S. = sin 2A = 2 sin A cos A
Multiply numerator and denominator by cos²A:
= (2 sin A cos A × cos²A) / cos²A
= (2 sin A/cos A) / (1/cos²A)
= (2 tan A) / (sec²A)
= 2 tan A / (1 + tan²A) = R.H.S. ✓ Proved
6
त्रिकोणमिति / Trigonometry — Heights & Distances
100 m अग्लो चट्टानको टुप्पोबाट एकै दिशामा रहेका जमिनका दुई बिन्दुहरू अवलोकन गर्दा क्रमशः अवनति कोणहरू 60° र 45° पाइयो ।
From the top of a cliff of height 100 m, two points on the ground in the same direction are observed. The angles of depression are 60° and 45° respectively.
a. चट्टानको उचाइ, जमिनका दुई बिन्दु र अवनति कोणहरू उल्लेख गरी चित्र बनाउनुहोस् । b. जमिनका दुई बिन्दुबिचको दूरी पत्ता लगाउनुहोस् ।
1+2
✓ Model Answer
a. Diagram (1 mark): T is top of cliff, C is base, P is nearer point (angle 60°), Q is farther point (angle 45°). Height TC = 100 m, angles of depression from T to P and Q are 60° and 45°.
b. Finding distance PQ (2 marks):
From T, angle of depression to P = 60°: tan 60° = 100/CP → CP = 100/√3 m
From T, angle of depression to Q = 45°: tan 45° = 100/CQ → CQ = 100 m
अनुबन्धित सर्वसमिका दिइएको छ: cos 2A + cos 2B + cos 2C = −1 − 4 cos A cos B cos C
Given conditional identity: cos 2A + cos 2B + cos 2C = −1 − 4 cos A cos B cos C
a. 2 cos A cos B लाई योग वा अन्तरको रूपमा लेख्नुहोस् । b. यदि A = B = C = 60° भए माथि दिइएको सर्वसमिका प्रमाणित गर्नुहोस् । c. A = 90°, B = 60° र C = 45° राख्दा दिइएको सर्वसमिका प्रमाणित नहुनुको सम्भावित कुनै एक कारण लेख्नुहोस् ।
1+2+1
✓ Model Answer
a. Product to Sum (1 mark): 2 cos A cos B = cos(A–B) + cos(A+B)
b. Verification for A=B=C=60° (2 marks):
L.H.S. = cos 2(60°) + cos 2(60°) + cos 2(60°) = cos120° + cos120° + cos120°
= (–1/2) + (–1/2) + (–1/2) = –3/2
c. Reason for A=90°, B=60°, C=45° (1 mark):
This conditional identity holds only when A + B + C = 180° (i.e., A, B, C are angles of a triangle).
Here: 90° + 60° + 45° = 195° ≠ 180°, so the condition is not satisfied and the identity cannot be verified.
सिधा रेखा L₁ बिन्दुहरू A(1, 2) र B(5, 6) भएर गएको छ ।
A straight line L₁ passes through the points A(1, 2) and B(5, 6).
a. बिन्दु A(1, 2) लाई बिन्दु A'(–2, –1) मा स्थानान्तरण गर्ने मेट्रिक्स लेख्नुहोस् । b. सिधा रेखा L₁ ले सिधा रेखा L₂: 2x – y + 3 = 0 सँग बनाउने न्यूनकोण θ छ भने tan θ को मान पत्ता लगाउनुहोस् ।
1+2
✓ Model Answer
a. Translation Matrix (1 mark):
A(1,2) → A'(–2,–1): shift = (–2–1, –1–2) = (–3, –3)
Translation vector/matrix: T = [–3, –3]ᵀ or the translation vector is (–3, –3).
b. Angle between lines (2 marks):
Slope of L₁: m₁ = (6–2)/(5–1) = 4/4 = 1
Slope of L₂ (2x–y+3=0 → y=2x+3): m₂ = 2
एउटा वृत्तको समीकरण x² + y² − 6x − 4y − 12 = 0 छ ।
x² + y² − 6x − 4y − 12 = 0 is an equation of a circle.
a. केन्द्र (h, k) र अर्धव्यास (r) भएको वृत्तको समीकरण लेख्नुहोस् । b. माथि दिइएको वृत्तको केन्द्र र अर्धव्यास पत्ता लगाउनुहोस् । c. उक्त वृत्तको परिधिको कुनै बिन्दु P(8, 2) छ भने दुई बिन्दुरूपको सूत्र प्रयोग गरी वृत्तको अर्धव्यासको समीकरण पत्ता लगाउनुहोस् ।
1+2+2
✓ Model Answer
a. Standard Equation (1 mark): (x – h)² + (y – k)² = r²
b. Centre and Radius (2 marks):
x² + y² – 6x – 4y – 12 = 0
Compare with x²+y²+2gx+2fy+c=0: g=–3, f=–2, c=–12
Centre = (–g, –f) = (3, 2)
r = √(g²+f²–c) = √(9+4+12) = √25 = 5 units
c. Equation of radius through P(8,2) (2 marks):
Centre C(3,2), Point P(8,2)
Using two-point formula: (y–y₁)/(y₂–y₁) = (x–x₁)/(x₂–x₁)
(y–2)/(2–2) = (x–3)/(8–3)
0 = (x–3)/5 × (y–2) — both points have same y-coordinate y=2 ∴ Equation of radius: y = 2
10
ज्यामिति / Geometry — Combined Transformation
एउटा सिधा रेखाले बिन्दुहरू A(2, 1) र B(6, 3) लाई जोडिएको छ ।
Points A(2, 1) and B(6, 3) are joined by a straight line.
a. दिइएका बिन्दुहरू A र B लाई संयुक्त स्थानान्तरण ToR ले स्थानान्तरण गर्दा प्राप्त हुने प्रतिबिम्ब बिन्दुहरू पत्ता लगाउनुहोस्, जहाँ R ले Y-अक्षमा भएको परावर्तन र T ले भेक्टर (3, 2) मा भएको विस्थापन बुझाउँछ । b. वस्तु र प्रतिबिम्ब दुवै चित्रलाई एउटै लेखाचित्रमा देखाउनुहोस् ।
2+1
✓ Model Answer
a. Combined Transformation ToR (2 marks): Step 1 — Reflection R in Y-axis (x→–x, y unchanged):
A(2,1) → A'(–2, 1)
B(6,3) → B'(–6, 3)
Step 2 — Translation T by vector (3, 2) (add 3 to x, 2 to y):
A'(–2,1) → A''(–2+3, 1+2) = A''(1, 3)
B'(–6,3) → B''(–6+3, 3+2) = B''(–3, 5)
A''(1, 3) and B''(–3, 5)
b. Graph (1 mark): Plot A(2,1), B(6,3) as object line and A''(1,3), B''(–3,5) as image line on the same axes. Label all four points.
11
भेक्टर / Vector
त्रिभुज ABC का शीर्षविन्दुहरूका निर्देशाङ्क A(2,3), B(4,1) र C(6,5) छन् । AB र AC का मध्यविन्दुहरू क्रमशः D र E छन् ।
The coordinates of vertices of triangle ABC are A(2,3), B(4,1) and C(6,5). D and E are midpoints of AB and AC respectively.
a. D को स्थिति भेक्टरलाई A र B का स्थिति भेक्टरका रूपमा लेख्नुहोस् । b. D को स्थिति भेक्टर i⃗ र j⃗ को रूपमा पत्ता लगाउनुहोस् । c. भेक्टर विधिबाट प्रमाणित गर्नुहोस्: DE⃗ = (1/2)BC⃗
1+1+1
✓ Model Answer
a. Position vector of D (1 mark): OD⃗ = (OA⃗ + OB⃗) / 2
b. D in terms of i⃗ and j⃗ (1 mark):
OA⃗ = 2i + 3j, OB⃗ = 4i + j
OD⃗ = (OA⃗+OB⃗)/2 = (6i+4j)/2 = 3i + 2j
तलको बक्स प्लटहरूमा ब्याट्री X र Y का टिकाउ क्षमता देखाइएको छ ।
The box plots show the battery life of Battery X and Battery Y. Battery Y median ≈ 82h; Battery X median ≈ 80h. Battery Y: Q₁≈76, Q₃≈88; Battery X: Q₁≈76, Q₃≈86.
a. मध्यिकाका आधारमा कुन ब्याट्रीको टिकाउ क्षमता उच्च छ ? b. ब्याट्री Y को चतुर्थांशीय विचलनको गुणाङ्क 0.098 छ भने ब्याट्री X को चतुर्थांशीय विचलनको गुणाङ्क निकाल्नुहोस् । c. एउटा टेलिकम कम्पनीलाई दुर्गममा रहेको signal tower मा पावरका लागि ब्याट्री X र Y मध्ये एउटा छनोट गर्नुपर्छ । माथिको सूचनाका आधारमा तपाईं कुन ब्याट्री प्रयोगका लागि सुझाव दिनुहुन्छ, किन ?
1+1+1
✓ Model Answer
a. Higher Median (1 mark):
From the box plots, Battery Y has the higher median battery life (median of Y ≈ 82h > median of X ≈ 80h).
b. CQD of Battery X (1 mark):
From box plot of X: Q₁ ≈ 76, Q₃ ≈ 86
CQD of X = (Q₃–Q₁)/(Q₃+Q₁) = (86–76)/(86+76) = 10/162 ≈ 0.0617
(Accept approximately 0.056–0.062 depending on reading of box plot values)
c. Recommendation (1 mark): Battery X is recommended for the remote signal tower.
Reason: A lower CQD (Coefficient of Quartile Deviation) indicates more consistent and reliable performance. Battery X has a lower CQD than Battery Y, meaning its battery life is more predictable and stable — critical for a remote tower where maintenance is difficult.
13
सीमान्तमान र निरन्तरता / Limit and Continuity
एउटा फलन p(x) निम्नानुसार परिभाषित गरिएको छ: p(x) = 2x+1 for 1≤x≤3 ; p(x) = x+4 for x>3
a. फलनको अविच्छिन्नतालाई परिभाषित गर्नुहोस् । b. x को मान 3 को नजिक हुँदा फलन p(x) को सीमान्तमान पत्ता लगाउनुहोस् । c. p(3) पत्ता लगाउनुहोस् । d. x = 3 मा फलन p(x) अविच्छिन्न वा विच्छिन्न के छ पत्ता लगाउनुहोस् र आफ्नो निष्कर्षलाई पुष्टि गर्नुहोस् ।
1+1+1+1
✓ Model Answer
a. Definition of Continuity (1 mark):
A function f(x) is continuous at x = a if: (1) f(a) is defined, (2) lim(x→a) f(x) exists (LHL = RHL), and (3) lim(x→a) f(x) = f(a).
b. Limit as x→3 (1 mark):
LHL (x→3⁻, using 2x+1): lim = 2(3)+1 = 7
RHL (x→3⁺, using x+4): lim = 3+4 = 7
LHL = RHL = 7, so lim(x→3) p(x) = 7
c. p(3) (1 mark):
Since 3 is in the interval 1≤x≤3, use p(x) = 2x+1:
p(3) = 2(3)+1 = 7
d. Conclusion (1 mark):
Check all three conditions: (1) p(3) = 7 ✓ defined, (2) limit exists = 7 ✓, (3) limit = p(3) = 7 ✓
Since all three conditions are satisfied:
∴ p(x) is CONTINUOUS at x = 3 ✓
समूह 'ग' — अन्तरविषयक्षेत्रगत प्रश्न (Group C — Cross Content Area)
Questions 14–17: integrating multiple content areas
[4×6 = 24]
14
CCA — Geometry + Algebra (Matrix)
शीर्षविन्दुहरू O(0,0), A(1,0) र B(1,1) भएको त्रिभुज OAB र एउटा स्थानान्तरण मेट्रिक्स M = [[0,1],[1,2]] छ ।
There is triangle OAB with vertices O(0,0), A(1,0) and B(1,1), and transformation matrix M = [[0,1],[1,2]].
a. त्रिभुज OAB लाई मेट्रिक्स M ले स्थानान्तरण गर्दा बन्ने प्रतिबिम्ब त्रिभुजका शीर्षविन्दुहरू O'A'B' का निर्देशाङ्क पत्ता लगाउनुहोस् । b. त्रिभुज OAB मा शीर्षविन्दु O बाट AB मा खिचेको उचाइको झुकाव पत्ता लगाउनुहोस् । c. के M एकल मेट्रिक्स हो ? यदि होइन भने कसरी एकल मेट्रिक्स बनाउनुहुन्छ ? d. Dolma को भनाइ छ मेट्रिक्स M वा यसको क्रम परिवर्तन गर्दा बन्ने मेट्रिक्स Mᵀ ले गर्ने स्थानान्तरण समान हुन्छ । के तपाईं Dolma को भनाइसँग सहमत हुनुहुन्छ, उदाहरणसहित पुष्टि गर्नुहोस् ।
2+1+1+2
✓ Model Answer
a. Image vertices (2 marks):
M = [[0,1],[1,2]], Object matrix = [[0,1,1],[0,0,1]]
b. Slope of altitude from O to AB (1 mark):
A=(1,0), B=(1,1) → AB is the vertical line x=1 (slope undefined = ∞)
Altitude from O perpendicular to AB must be horizontal: slope = 0
c. Singular matrix check (1 mark):
det(M) = (0)(2) – (1)(1) = –1 ≠ 0
So M is NOT a singular matrix. To make it singular, change one element so det=0. e.g. change M[0][1] from 1 to 0: [[0,0],[1,2]] → det=0 ✓
d. Dolma's claim (2 marks):
Mᵀ = [[0,1],[1,2]] — Note: For this particular matrix M = [[0,1],[1,2]], the transpose Mᵀ = [[0,1],[1,2]] = M (since M is symmetric).
Therefore M = Mᵀ, and both produce identical transformations. Dolma's claim is CORRECT for this specific matrix. However, this is only because M happens to be symmetric — in general, M and Mᵀ produce different transformations.
15
CCA — Geometry + Vector (Conic Sections + Circle)
Shreya ले एउटा काठको सोलीलाई साङ्किक क्षेत्रको स्वरूपमा काट्नुभयो र सोलीको साङ्किक क्षेत्रका किनाराहरू कोर्नुभयो ।
Shreya cuts a wooden cone to form a conic section and draws its outline on paper.
a. सोलीलाई एउटा समतलीय सतहले semi vertical angle भन्दा ठूलो र 90° भन्दा सानो हुने गरी छड्के काट्दा कागजमा बन्ने आकारको नाम लेख्नुहोस् । b. Shreya ले अर्धवृत्त ABC बनाउनुभयो र अर्धवृत्तको परिधिको बिन्दु B बाट ब्यासका दुई छेउ A र C मा खिचेका जीवाहरू सधैं समकोण बनाउँछन् भन्ने पाउनुभयो । यो भनाइलाई भेक्टर विधिबाट प्रमाणित गर्नुहोस् । c. के Shreya को भनाइ बिन्दु B अर्धवृत्तभित्र चापमा पर्दा वा नपर्दा पनि सही हुन्छ, कारण लेख्नुहोस् । d. यदि (AB)² = √x + 9 र (BC)² = √x र AC = 3 एकाइ भए x को मान निकाल्नुहोस् ।
1+2+1+2
✓ Model Answer
a. Conic section name (1 mark):
When the cutting plane makes an angle greater than the semi-vertical angle and less than 90° with the axis, the conic section formed is a HYPERBOLA.
b. Angle in semicircle = 90° — Vector Proof (2 marks):
Let O be the centre of the semicircle with radius r. Let position vectors: OA⃗ = a⃗, OB⃗ = b⃗, OC⃗ = –a⃗ (since AC is diameter, C is opposite to A).
BA⃗ = OA⃗ – OB⃗ = a⃗ – b⃗
BC⃗ = OC⃗ – OB⃗ = –a⃗ – b⃗
BA⃗ · BC⃗ = (a⃗–b⃗)·(–a⃗–b⃗) = –|a⃗|² – a⃗·b⃗ + b⃗·a⃗ + |b⃗|² = |b⃗|² – |a⃗|²
Since B is on the circle: |OB⃗| = |OA⃗| = r, so |b⃗| = |a⃗|
∴ BA⃗ · BC⃗ = r² – r² = 0 → BA ⊥ BC ✓ Proved
c. B inside the semicircle (1 mark): No — Shreya's claim does NOT hold if B is inside the semicircle.
If B is inside the circle, |OB⃗| < r, so |b⃗| < |a⃗|, giving BA⃗·BC⃗ = |b⃗|²–|a⃗|² < 0. A negative dot product means the angle ∠ABC is obtuse (greater than 90°), not a right angle.
d. Finding x (2 marks):
Since angle in semicircle = 90°, Pythagoras applies: AB² + BC² = AC²
(√x + 9) + √x = 3² = 9
2√x + 9 = 9
2√x = 0 → √x = 0 → x = 0
If the question means (AB)² = √(x+9) and (BC)² = √x:
√(x+9) + √x = 9
Let √x = t: √(t²+9) = 9–t → squaring: t²+9 = 81–18t+t² → 18t=72 → t=4 → √x=4 → x=16
x = 16
16
CCA — Trigonometry + Algebra (Functions + Inverse)
एक जना पदयात्री एउटा पहाड चढिरहेको छ । उक्त पहाडको उचाइ र क्षितिजीय दूरी मिटरमा क्रमशः h र x सँग सम्बन्धित फलन h = f(x) = mx + 50 छ । पहाडको झुकाव m = (cosθ – cos3θ)/(sin3θ – sinθ) छ ।
A hiker is climbing a hill. h = f(x) = mx + 50 (height in terms of horizontal distance). Slope m = (cosθ – cos3θ)/(sin3θ – sinθ).
a. प्रमाणित गर्नुहोस्: m = tan 2θ b. पहाडको कोण θ यस्तो छ कि tan θ = 1/3 भए m को मान पत्ता लगाउनुहोस् । c. विपरीत फलन f⁻¹(x) को प्रयोग गरी पदयात्री 150 मिटर उचाइमा हुँदा पार गरेको क्षितिजीय दूरी पत्ता लगाउनुहोस् । d. Bina को भनाइ: "यदि कुनै रेखाले X-अक्षको धनात्मक दिशामा बनाएको कोण दुई गुणा भएमा सो रेखाको झुकाव पनि दुई गुणा हुन्छ ।" tan 2θ र θ को कुनै एक मानको प्रयोग गरी वहाँको भनाइ पुष्टि गर्नुहोस् ।
2+2+2+1
✓ Model Answer
a. Prove m = tan 2θ (2 marks):
m = (cosθ – cos3θ) / (sin3θ – sinθ)
Numerator: cosθ – cos3θ = 2 sin2θ · sinθ [sum-to-product: cos A – cos B = –2sin((A+B)/2)sin((A–B)/2)]
Denominator: sin3θ – sinθ = 2 cos2θ · sinθ [sin A – sin B = 2cos((A+B)/2)sin((A–B)/2)]
m = (2 sin2θ · sinθ) / (2 cos2θ · sinθ) = sin2θ/cos2θ = tan 2θ ✓ Proved
b. Value of m when tan θ = 1/3 (2 marks):
m = tan 2θ = 2 tan θ / (1 – tan²θ) = 2(1/3) / (1 – 1/9) = (2/3) / (8/9) = (2/3) × (9/8) = 3/4
m = 3/4
c. Horizontal distance at h = 150m (2 marks):
f(x) = mx + 50 = (3/4)x + 50
Find f⁻¹(x): y = (3/4)x + 50 → x = (y–50) × 4/3 = 4(y–50)/3
f⁻¹(150) = 4(150–50)/3 = 4(100)/3 = 400/3 ≈ 133.3 m
d. Bina's Claim (1 mark):
Bina's claim is FALSE. Example: Let θ = 45°, so slope = tan 45° = 1. If angle doubles to 90°, slope = tan 90° = undefined (∞), not 2. The relationship between angle and slope is nonlinear (tangent function), not linear doubling.
17
CCA — Statistics + Limit + Algebra
सडक विभागले एउटा सिधा सडकको k मिटरमा एउटा बस स्टप राख्न चाहन्छ ।
Department of Road wants to place a bus stop at position k metres along a straight road. Residents per section: 0–200: 20, 200–400: 30, 400–600: 40, 600–800: 10. The total squared walking distance is s(k) = 10k² – 76k + 178.
a. सडक छेउमा रहेका बासिन्दाको वितरणको विचरणशीलताको गुणाङ्क निकाल्नुहोस् । b. के k का सबै मानका लागि फलन s(k) एक अविच्छिन्न फलन हो ? c. Anu ले हिसाब गर्दा s(3.8) = 30 पाउनुभयो । के यस मानले 3.8 मा फलन s(k) अविच्छिन्न हुन्छ ?
3+1+1
✓ Model Answer
a. Coefficient of Variation (CV) (3 marks):
Data: mid-values m = 100, 300, 500, 700; frequencies f = 20, 30, 40, 10; N = 100
b. Is s(k) continuous? (1 mark): Yes. s(k) = 10k² – 76k + 178 is a polynomial function. All polynomial functions are continuous for every real value of their variable — there are no breaks, holes, or asymptotes.
c. Anu's value s(3.8) = 30 — is s continuous at 3.8? (1 mark):
Actual value: s(3.8) = 10(3.8)² – 76(3.8) + 178 = 10(14.44) – 288.8 + 178 = 144.4 – 288.8 + 178 = 33.6
Anu's calculated value (30) is incorrect. The arithmetic error does not affect the continuity of the function — s(k) is still continuous at 3.8 because it is a polynomial. A calculation mistake does not change the nature of the function.
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📊 Summary — Optional Mathematics New Curriculum Model Q 2083